> #optics , #physics > A microscope objective lens has a diameter of 6.8mm and a focal length of 4.0 mm ,for a sample viewed in air what is the resolving power of this objective in ; a) red. b)blue lights?
Anonymous
more than 3 weeks ago

A microscope objective lens has a diameter of 6.8mm and a focal length of 4.0 mm ,for a sample viewed in air what is the resolving power of this objective in ; a) red. b)blue lights?

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ChatGPT
more than 3 weeks ago
The resolving power of a microscope objective can be calculated using the formula:

Resolving power = 0.61 * (wavelength of light) / (numerical aperture)

a) For red light, the wavelength is approximately 700 nm. The numerical aperture of the objective can be calculated using the formula:

Numerical aperture = (n * sin(θ))

where n is the refractive index of the medium (1 for air) and θ is half the angle of the cone of light entering the objective. The angle can be calculated using the formula:

sin(θ) = (diameter of objective / (2 * focal length))

Plugging in the values:

sin(θ) = (6.8 mm / (2 * 4.0 mm)) = 0.85

Numerical aperture = 1 * 0.85 = 0.85

Now, plug in the values into the resolving power formula:

Resolving power = 0.61 * (700 nm) / 0.85 = 504 nm

Therefore, the resolving power of the objective in red light is 504 nm.

b) For blue light, the wavelength is approximately 450 nm. Using the same numerical aperture calculated above, plug in the values into the resolving power formula:

Resolving power = 0.61 * (450 nm) / 0.85 = 324 nm

Therefore, the resolving power of the objective in blue light is 324 nm.
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Google Gemini AI
more than 3 weeks ago
The resolving power of a microscope objective lens is given by:

$$d = 1.22 \frac{\lambda}{2NA}$$

where:

- d is the resolving power
- λ is the wavelength of light
- NA is the numerical aperture of the objective lens

The numerical aperture of an objective lens is given by:

$$NA = n\sin\theta$$

where:

- n is the index of refraction of the medium between the objective lens and the sample
- θ is the half-angle of the cone of light that enters the objective lens

For a sample viewed in air, the index of refraction is 1.00. The half-angle of the cone of light that enters the objective lens can be determined from the focal length and the diameter of the objective lens:

$$\sin\theta = \frac{d/2}{f}$$

where:

- d is the diameter of the objective lens
- f is the focal length of the objective lens

Therefore, the numerical aperture of the objective lens is:

$$NA = 1.00 \sin\theta = 1.00 \frac{d/2}{f} = 1.00 \frac{6.8\text{ mm}/2}{4.0\text{ mm}} = 0.85$$

a) For red light, the wavelength is approximately 650 nm. Therefore, the resolving power of the objective lens is:

$$d = 1.22 \frac{\lambda}{2NA} = 1.22 \frac{650\text{ nm}}{2 \times 0.85} = 468\text{ nm}$$

b) For blue light, the wavelength is approximately 450 nm. Therefore, the resolving power of the objective lens is:

$$d = 1.22 \frac{\lambda}{2NA} = 1.22 \frac{450\text{ nm}}{2 \times 0.85} = 328\text{ nm}$$
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